Math, asked by Anonymous, 10 months ago

1. Prove that
 \sqrt{5}  -  \sqrt{3}
is not a rational number.​

Answers

Answered by Anonymous
2

Answer

Let \tt{\sqrt{5} - \sqrt{3} } be a rational number.

\tt{\therefore \: \sqrt{5} - \sqrt{3} = \frac{p}{q}}\\, where q and p are co-primes with no common factor other than 1

Squaring on both sides:

\tt{(\sqrt{5} - \sqrt{3})^{2} = \frac{p^{2}}{q^{2}}}\\

We get:

\tt{(\sqrt{5} - \sqrt{3})^{2}q^{2} = p^{2}}\\ ---(1)

Thus;

\tt{(\sqrt{5} - \sqrt{3})^{2}} || \tt{p^{2}}

And so,

\tt{(\sqrt{5} - \sqrt{3})^{2}} || p

Now, we know that:

p = \tt{(\sqrt{5} - \sqrt{3})^{2} m}

Squaring on both sides:

\tt{p^{2} = (\sqrt{5} - \sqrt{3})^{4} m^{2}}

Put (1):

\tt{(\sqrt{5} - \sqrt{3})^{2} q^{2} = (\sqrt{5} - \sqrt{3})^{4}m^{2} }

Crossing the same terms, we get:

\tt{(\sqrt{5} - \sqrt{3})^{2}m^{2} = q^{2}}

Thus;

\tt{(\sqrt{5} - \sqrt{3})^{2}} || \tt{q^{2}}

And so,

\tt{(\sqrt{5} - \sqrt{3})^{2}} || q

But p and q were co-primes.

This contradicts our assumption. As, p and q have common factor as \tt{(\sqrt{5} - \sqrt{3})^{2}}.

Thus, \tt{\sqrt{5} - \sqrt{3} } is an irrational number.

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